What is the range of the quadratic function?

What represents the inverse of the function f(x) = 4x

Katarina [22]

Answer:

$f { - 1}^{ } = \frac{x}{4}$

Step-by-step explanation:

fjhjthfdvhfjrhgehdhrjfhdrhjthr

8 0

3 years ago

Rotate the point (1,2) 90° (-4,4) (0,2) (2,0) (-2,1)

saw5 [17]

Answer: (-2,1)

Step-by-step explanation:

The point (1,2) is in quadrant 1.

When rotated 90 degrees counter clockwise, it moves to quadrant 2.

The new point becomes (-2,1)

5 0

3 years ago

(b²) ____ = b⁻⁶ find the missing exponent

allsm [11]

$(b^2)^{-3}=b^{2*(-3)}=b^{-6}$

the missing exponent is -3

6 0

3 years ago

Slope for y=5,520.619x - 1,091.393

Bad White [126]

Answer:

m = 5,520.619

Step-by-step explanation:

The given equation is :

y=5,520.619x-1,091.393 ....(1)

The general form of equation is given by :

y = mx +c ...(2)

Where

m is slope of line

c is y-intercept

Comparing equation (1) and (2), we get :

Slope, m = 5,520.619

Hence, the slope of the given line is 5,520.619.

3 0

3 years ago

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be

irga5000 [103]

Answer:

The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

Step-by-step explanation:

Let $C = (h,k)$ the coordinates of the center of the circle, which must be transformed into $C'=(h', k')$ by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

$(x-h)^{2}+(y-k)^{2} = r^{2}$

Where $r$ is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) Center of the circle is translated 4 units to the right (+x direction):

$C''(x,y) = C(x, y) + U(x,y)$ (Eq. 1)

Where $U(x,y)$ is the translation vector, dimensionless.

If we know that $C(x, y) = (h,k)$ and $U(x,y) = (4, 0)$ , then:

$C''(x,y) = (h,k)+(4,0)$

$C''(x,y) =(h+4,k)$

2) Reflection over the x-axis:

$C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]$ (Eq. 2)

Where $O(x,y)$ is the reflection point, dimensionless.

If we know that $O(x,y) = (h+4,0)$ and $C''(x,y) =(h+4,k)$ , the new point is:

$C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]$

$C'(x,y) = (h+4, 0)-(0,k)$

$C'(x,y) = (h+4, -k)$

And thus, $h' = h+4$ and $k' = -k$ . The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

4 0

3 years ago